Grade 12th passMechanics

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Rahul

9 Years agoGrade 12th pass

1 Answer

Kenny Toijam

9 Years ago

Since the two blocks are moving together as they are in contact.Therefore the acceleration of the whole system is equal to the acceleration of the 2Kg mass

Now to calculate this question,Draw the F.B.D of the system of blocks.

The forces acting on the 6kg mass are:

1.(M+m)?gsin30

2.f1+f2 i.e Net friction acting on system

Now apply friction equations and find net force on system

fi=4gcos30*0.3 f2=2gcos30*0.2

F(net)=(4+2)gsin30-(f1 +f2)

6a=6gsin30-8*(root over 3)

a=2.6m/s2.............which is the accn of the block 2kg too.

So the CORRECT OPTION IS (B)

HOPE IT HELPS......

Yours faithfully,

Kenny Toijam.

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pls refer to the attachmentpls refer to the attachmentpls refer to the attachmentpls refer to the attachmentpls refer to the attachmentpls refer to the attachment

Rahul

1 Answer

Kenny Toijam

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pls refer to the attachment
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